Let $f_1,f_2,\dots,f_n : \mathbb{D} \to \mathbb{C}$ be holomorphic functions and consider the polynomial
$$ w^n + f_1(z)w^{n-1} + \dots + f_n(z). $$
Suppose, I happen to know that
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For each $z$, the roots of the above polynomial are all in $\mathbb{D}$.
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For each $z$, one of the roots is an $n$-th root of $z$.
Given these two conditions, is it true that for each $z$, all the roots of the above polynomial are nothing but $n$-th roots of $z$?
Best Answer
It is true. Here's the proof. Let $\zeta=e^{\frac{2\pi i}{n}}$ be the $n$-th root of unity. Note that by the assumption, for each $z\in \mathbb{D}$, there is $k\in \{0,1,\ldots,n-1\}$ such that $$ p(z^n, z\zeta^k) = 0 $$ where $$ p(z,w) = w^n + f_1(z)w^{n-1} + \cdots + f_n(z). $$ Let $D_k$ be the set of all $z\in \mathbb{D}$ such that $p(z^n, z\zeta^k) = 0$. Then each $D_k$ is closed in $\mathbb{D}$ and we have $$ \mathbb{D} = \bigcup_{0\leq k\leq n-1} D_k. $$ Note that by pigeonhole principle, one of $D_k$ has $0$ as its limit point. Then Identity theorem implies that for some $k$, $$ p(z^n, z\zeta^k) \equiv 0\quad\cdots(*), $$ for all $z\in\mathbb{D}$. Note that change of variable $z\mapsto z\zeta^j$ yields $$ p(z^n, z\zeta^{k+j})=0,\quad\forall j, $$ and hence $$ p(z^n, z\zeta^{j})=0, \quad \forall j=0,1,\ldots, n-1. $$ It is saying that $p(z,w)=0$ for $\textbf{all}$ roots of $w^n = z$, giving us the desired result that $$ p(z,w) = w^n -z. $$